Nội dung Text: Đề thi Olympic sinh viên thế giới năm 2003

10th International Mathematical Competition for University Students
Cluj-Napoca, July 2003
Day 1
1. (a) Let a1 , a2 , . . . be a sequence of real numbers such that a1 = 1 and an+1 > 3 an for all n.
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Prove that the sequence
an
3 n−1
2
has a ﬁnite limit or tends to inﬁnity. (10 points)
(b) Prove that for all α > 1 there exists a sequence a1 , a2 , . . . with the same properties such
that
an
lim n−1 = α.
3
2
(10 points)
an 3
Solution. (a) Let bn = . Then an+1 > 2 an is equivalent to bn+1 > bn , thus the sequence
3 n−1
2
(bn ) is strictly increasing. Each increasing sequence has a ﬁnite limit or tends to inﬁnity.
(b) For all α > 1 there exists a sequence 1 = b1 < b2 < . . . which converges to α. Choosing
3 n−1
an = 2 bn , we obtain the required sequence (an ).
2. Let a1 , a2 . . . , a51 be non-zero elements of a ﬁeld. We simultaneously replace each element with
the sum of the 50 remaining ones. In this way we get a sequence b1 . . . , b51 . If this new sequence is
a permutation of the original one, what can be the characteristic of the ﬁeld? (The characteristic
of a ﬁeld is p, if p is the smallest positive integer such that x + x + . . . + x = 0 for any element x
p
of the ﬁeld. If there exists no such p, the characteristic is 0.) (20 points)
Solution. Let S = a1 + a2 + · · · + a51 . Then b1 + b2 + · · · + b51 = 50S. Since b1 , b2 , · · · , b51 is a
permutation of a1 , a2 , · · · , a51 , we get 50S = S, so 49S = 0. Assume that the characteristic of the
ﬁeld is not equal to 7. Then 49S = 0 implies that S = 0. Therefore bi = −ai for i = 1, 2, · · · , 51.
On the other hand, bi = aϕ(i) , where ϕ ∈ S51 . Therefore, if the characteristic is not 2, the sequence
a1 , a2 , · · · , a51 can be partitioned into pairs {ai , aϕ(i) } of additive inverses. But this is impossible,
since 51 is an odd number. It follows that the characteristic of the ﬁeld is 7 or 2.
The characteristic can be either 2 or 7. For the case of 7, x1 = . . . = x51 = 1 is a possible
choice. For the case of 2, any elements can be chosen such that S = 0, since then bi = −ai = ai .
3. Let A be an n × n real matrix such that 3A3 = A2 + A + I (I is the identity matrix). Show
that the sequence Ak converges to an idempotent matrix. (A matrix B is called idempotent if
B 2 = B.) (20 points)
Solution. The minimal polynomial of A is a divisor of 3x3 − x2 − x − 1. This polynomial has three
diﬀerent roots. This implies that A is diagonalizable: A = C −1 DC where D is a diagonal matrix.
The eigenvalues of the matrices A and D are all roots of polynomial 3x3 − x2 − x − 1. One of the
three roots is 1, the remaining two roots have smaller absolute value than 1. Hence, the diagonal
elements of Dk , which are the kth powers of the eigenvalues, tend to either 0 or 1 and the limit
M = lim Dk is idempotent. Then lim Ak = C −1 M C is idempotent as well.
4. Determine the set of all pairs (a, b) of positive integers for which the set of positive integers
can be decomposed into two sets A and B such that a · A = b · B. (20 points)
Solution. Clearly a and b must be diﬀerent since A and B are disjoint.
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Denote by Br (a) = {x ∈ Rn : |x − a| ≤ r} and ∂Br (a) = {x ∈ Rn : |x − a| = r} the
ball and the sphere of center a and radius r, respectively.
If a0 is not the unique nearest point then for any point a on the open line segment (a0 , b0 )
we have B|a−a0 | (a) ⊂ B (b0 ) and ∂B|a−a0 | (a) ∂B (b0 ) = {a0 }, therefore (a0 , b0 ) ⊂ B and
b0 is an accumulation point of set B.
4. Find all positive integers n for which there exists a family F of three-element subsets
of S = {1, 2, . . . , n} satisfying the following two conditions:
(i) for any two diﬀerent elements a, b ∈ S, there exists exactly one A ∈ F containing
both a, b;
(ii) if a, b, c, x, y, z are elements of S such that if {a, b, x}, {a, c, y}, {b, c, z} ∈ F, then
{x, y, z} ∈ F.
Solution. The condition (i) of the problem allows us to deﬁne a (well-deﬁned) operation
∗ on the set S given by
a ∗ b = c if and only if {a, b, c} ∈ F, where a = b.
We note that this operation is still not deﬁned completely (we need to deﬁne a ∗ a), but
nevertheless let us investigate its features. At ﬁrst, due to (i), for a = b the operation
obviously satisﬁes the following three conditions:
(a) a = a ∗ b = b;
(b) a ∗ b = b ∗ a;
(c) a ∗ (a ∗ b) = b.
What does the condition (ii) give? It claims that
(e’) x ∗ (a ∗ c) = x ∗ y = z = b ∗ c = (x ∗ a) ∗ c
for any three diﬀerent x, a, c, i.e. that the operation is associative if the arguments are
diﬀerent. Now we can complete the deﬁnition of ∗. In order to save associativity for non-
diﬀerent arguments, i.e. to make b = a ∗ (a ∗ b) = (a ∗ a) ∗ b hold, we will add to S an extra
element, call it 0, and deﬁne
(d) a ∗ a = 0 and a ∗ 0 = 0 ∗ a = a.
Now it is easy to check that, for any a, b, c ∈ S ∪ {0}, (a),(b),(c) and (d), still hold, and
(e) a ∗ b ∗ c := (a ∗ b) ∗ c = a ∗ (b ∗ c).
We have thus obtained that (S ∪ {0}, ∗) has the structure of a ﬁnite Abelian group,
whose elements are all of order two. Since the order of every such group is a power of 2,
we conclude that |S ∪ {0}| = n + 1 = 2m and n = 2m − 1 for some integer m ≥ 1.
Given n = 2m −1, according to what we have proven till now, we will construct a family
of three-element subsets of S satisfying (i) and (ii). Let us deﬁne the operation ∗ in the
following manner:
if a = a0 + 2a1 + . . . + 2m−1 am−1 and b = b0 + 2b1 + . . . + 2m−1 bm−1 , where ai , bi
are either 0 or 1, we put a ∗ b = |a0 − b0 | + 2|a1 − b1 | + . . . + 2m−1 |am−1 − bm−1 |.
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